Answer
a) $m=-1$
b) $y=-x+1$
Work Step by Step
We have: $x^2y-y^2+x=1$
We differentiate both sides with respect to $x$.
$x^2 \dfrac{dy}{dx}-2y \dfrac{dy}{dx}=-1 -2xy\\ \dfrac{dy}{dx}=\dfrac{-1-2xy}{x^2-2y} $
a) The slope at $( 1, 0)$ is: $m=\dfrac{-1-2(1)(0)}{1-2(0)} = -1$
b) The equation of a tangent line at $( 1,0)$ is:
$y-y_1=m(x-x_1)\\ y -0=-1(x-1)\\ y=-x+1$
